# How do you graph 2x+y<=4 on the coordinate plane?

Aug 25, 2017

See a solution process below:

#### Explanation:

First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For $x = 0$

$\left(2 \cdot 0\right) + y = 4$

$0 + y = 4$

$y = 4$ or $\left(0 , 4\right)$

For $y = 0$

$2 x + 0 = 4$

$2 x = 4$

$\frac{2 x}{\textcolor{red}{2}} = \frac{4}{\textcolor{red}{2}}$

$x = 2$ or $\left(2 , 0\right)$

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.
The boundary line will be solid because the inequality operator contains a "or equal to" clause.

graph{(x^2+(y-4)^2-0.075)((x-2)^2+y^2-0.075)(2x+y-4)=0 [-15, 15, -7.5, 7.5]}

Now, we can shade the left side of the line for the "less than" clause in the inequality.

graph{(2x+y-4)<=0 [-15, 15, -7.5, 7.5]}